$\begingroup$Being "locally affine" seems like the best possible answer: Being affine is an intrinsic property: $(X,O_X)$ is affine if morphisms from another locally ringed space to $X$ are the same as morphism of rings from $O_X$ to the global section of the other ringed space. And a scheme is exactly a locally affine locally ringed space. (to put it another way, you might want to precise what you mean by 'intrinsic')$\endgroup$
– Simon HenryOct 13 '16 at 9:26

$\begingroup$@SimonHenry: Well, "locally affine" is a reformulation of the definition; this is not what I'm looking for. I am looking for "global" properties. By "intrinsic" I meant that one doesn't just say that the topos is spatial etc. Sorry for being so vague here. I could also ask the question as follows: Assume that you have constructed a locally ringed topos (and the construction is sufficiently complicated), and you want to know if it comes from a scheme. Is there any procedure for this?$\endgroup$
– HeinrichDOct 13 '16 at 9:50

$\begingroup$I understand, but as said Ingo there is no known simple characterization, and on the other hand the definition as "locally affine" is in practice relatively easy to investigate: it is in general relatively easy to see whether a ringed space is affine or not, you just need to see if it is isomorphic to the spectrum of its global section, so you need to figure out if your ringed space has some affine open subspace...$\endgroup$
– Simon HenryOct 13 '16 at 10:24

1 Answer
1

Unfortunately I don't know an interesting intrinsically formulated sufficient criterion for a locally ringed topos to be the little Zariski topos of a scheme. This is an extremely interesting question!

There are necessary conditions, for instance (formulated in the internal language of the topos):

For any element $f : \mathcal{O}_X$: If $f$ is not invertible, then $f$ is nilpotent.

While this condition does exclude some locally ringed toposes, it doesn't exclude the locally ringed topos given by a smooth manifold (if a smooth function has the property that the only open subset on which it is (multiplicately) invertible is the empty set, then it is zero).

On a reduced scheme, where it holds that $\mathcal{O}_X$ is internally a reduced ring, a consequence of this condition is that $\mathcal{O}_X$ is "$\neg\neg$-separated": For any $f : \mathcal{O}_X$, if $\neg\neg(f = 0)$, then in fact $f = 0$.

Update: There is a further, but more convoluted, property which is enjoyed by the little Zariski topos of any scheme but not in general by the topos given by a smooth manifold:

For any element $f : \mathcal{O}_X$ the localized module $\mathcal{O}_X[f^{-1}]$ is a sheaf with respect to the internal modality $\Box$, where $\Box\varphi :\equiv (\text{$f$ invertible} \Rightarrow \varphi)$.

This condition can be rephrased so as to not refer to internal modalities:

For any element $f : \mathcal{O}_X$ it holds that:

For any $s : \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow s = 0$, there is a natural number $n$ such that $f^n s = 0$.

For any subset $K \subseteq \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow \text{$K$ is a singleton}$, there is a natural number $n$ and an element $s : \mathcal{O}_X$ such that $\text{$f$ invertible} \Rightarrow f^{-n} s \in K$.

The first part is satisfied by (the topos of sheaves over a) manifold, but the second part is not. A counterexample is $X = \mathbb{R}^1$, $f(x) = x$, $K = \text{"$e^{1/x}$ on $x \neq 0$"}$. In fact a manifold satisfies this condition if and only if it is empty.

The big Zariski topos of a scheme $X$, that is approximately the topos of sheaves on the Zariski site $\mathrm{Sch}/X$ with structure sheaf $\mathbb{A}^1 : T \mapsto \mathcal{O}_T(T)$, has the following special properties, which are more helpful in distinguishing it from other locally ringed toposes:

For any element $f : \mathbb{A}^1$: If $f$ is not zero, then $f$ is invertible.

For any finitely presented $\mathbb{A}^1$-algebra $A$, the canonical map $A \to \mathrm{Hom}(\mathrm{Hom}_{\mathbb{A}^1\mathrm{-Alg}}(A, \mathbb{A}^1), \mathbb{A}^1)$ is an isomorphism of $\mathbb{A}^1$-algebras.

In the special case that $A = \mathbb{A}^1[T]$, the second condition yields

Any map $\mathbb{A}^1 \to \mathbb{A}^1$ is given by a unique polynomial in $\mathbb{A}^1[T]$.

which somehow expresses the vague intution that "in algebraic geometry, every morphism is a polynomial".

$\begingroup$Thank you. A consequence of your condition is that if $\mathcal{O}_X$ is reduced, then $\mathcal{O}_X$ is actually a residue field (Elephant, D4.7), i.e. non-zero and all non-invertibles are zero. Actually I already knew this property (by reading your topos notes), but I wonder if there are other properties of the petit Zariski topos.$\endgroup$
– HeinrichDOct 13 '16 at 9:55

$\begingroup$@HeinrichD: Indeed. And this field property has many consequences, for instance generic freeness. I just added a further property to the answer, one which is sufficiently strong to distinguish schemes from manifolds.$\endgroup$
– Ingo BlechschmidtOct 24 '16 at 10:45

$\begingroup$Can you say more about 2. - in particular what it means concretely, why it is satisfied for affine schemes, and what do you mean by $K=$"$e^{1/x}$ on $x \neq 0$"? I thought that $K$ is a set of global sections.$\endgroup$
– HeinrichDOct 31 '16 at 17:45

$\begingroup$@HeinrichD: I'm sorry for the late reply. $K$ is only a set of global sections from the point of view of the internal language. From the external point, $K$ is a subsheaf of $\mathcal{O}_X$. The internal condition "$f$ is invertible $\Rightarrow$ $K$ is a singleton" means that on the open subset $D(f)$, the sheaf $K$ contains only a single section. The gist of the statement is that any such section can be extended to a global one, after clearing denominators. It is satisfied for affine schemes since the maps $\Gamma(X,\mathcal{O}_X)[f^{-1}] \to \Gamma(D(f),\mathcal{O}_X)$ are bijective.$\endgroup$
– Ingo BlechschmidtNov 11 '16 at 7:38

$\begingroup$@HeinrichD: By "$K = e^{1/x}$ on $x \neq 0$" I mean the subsheaf of the sheaf of continuous functions whose sections on an open subset $U\subseteq\mathbb{R}^1$ are given by $\{e^{1/x}\}$ if $0 \not\in U$ and $\emptyset$ otherwise. The reason why the condition fails for this choice of $K$ is that $e^{1/x}$, and also no product of $e^{1/x}$ with an arbitrarily high power of $x$, can be extended to a global continuous function. You can read some details (about the situation with schemes, not with manifolds) in Section 9 of these notes.$\endgroup$
– Ingo BlechschmidtNov 11 '16 at 7:49